COmbinations

Nine persons gather to play football by forming two teams of four to play each
other, the remaining person acting as referee.In how many different ways can the
teams be formed?
If two particular persons are not to be in the same team, how many ways are there
then to choose the teams?

Re: COmbinations

this is a 4unit hsc question (1981).

but, it could feasibly pop up, i guess? a question similar to this is in the New Senior Maths (fitzpatrick) book - q 30, chapter 3.4 - (In how many ways can eight basketball players be divided into four groups of two?)

anyways, SupR is right - you divide by 2, or i guess 2! (which is 2).

but the second part of SupR's solution doesn't make sense to me :S I'd do it by cases - A in group 1, A in group 2, A is ref.
Can someone else give their viewpoint? I'm probably wrong lel.

Re: COmbinations

If two particular persons are not to be in the same team, how many ways are there then to choose the teams?

actually, was thinking about this last night, how about:
First way to solve:
3 cases-
Case 1: B is ref - 8C4/2 <-- self explanatory.
Case 2: A is ref - 8C4/2 <-- self explanatory.
Case 3: A and B are both picked but on separate teams - 7C3 x 4C3 <-- A is already on first team, B is already on second team. pick people for remaining slot from the pool of 7, then repeat for the second team. don't divide by 2 here because A and B make the teams distinct.

Therefore, adding all 3 cases: 8C4/2 + 8C4/2 + (7C3 x 4C3) = 210

Second way to solve:
alternatively, total ways to pick minus both on same team:
(( 9C4 x 5C4 ) / 2) - ((7C2 x 5C4)) = 210
my reasoning here is as follows: 2 of them are definitely on the same team, so they aren't part of the choosing for the first team. 2 places remain, hence choose 2 from the remaining 7. then, for the second team, 5 choose 4. don't divide by 2 since teams distinguishable as a result of the 2 chosen.

Can someone please double check this? I'm really confused, but I think that's the right way to do it?

Re: COmbinations

still not 100% satisfied, so one last attempt:
A B C D
E F G H
I

Let's say those are the 9 guys, with A and B not to be on the same team.

Let's consider 9 cases, where each of 9 will be refs, then add all these 9 cases at the end. The reason i'm attempting this is because the 9th guy is the one that's making this problem confusing. if I can separate the problem into 9 cases, i can convert the problem into 9 separate 8C4 problems, which are super straight forward. 8 choose 4, and the left over 4 are the second team, so i don't even have to worry about the second choice. another way to think of it is 8C4 x 4c4.

Case 1: A is ref: 8C4/2! (or 8C4 x 4C4 / 2!) <-- same thing, depends on how you want to think of it.
Case 2: B is ref: 8C4/2! (or 8C4 x 4C4 / 2!) <-- same thing, depends on how you want to think of it.

we divide by 2! in the above 2 cases because if you choose 4 such that: B C D E are on the first team, the remaining 4 (F G H I) are on the other team. BUT choosing F G H I for the first team will give the exact same result - ie B C D E on one team, F G H I in the other. Since order doesn't matter, this is double counting. Exactly half of all possible choices will be duplicates, so we divide by 2.

Cases 3-9: C-I are refs (7 cases): A and B will be on separate teams. therefore: 6C3 (or 6C3 x 3C3 - depends on how you want to think of it)
7 such cases, so 7 x 6C3

we DON'T divide by 2! here because A and B are locked in, and since A and B are unique, choosing A C D E and A F G H (with i as ref, for example) are two separate cases.

answer: 8C4/2! + 8C4/2! + 7 x 6C3 = 210

i'm pretty sure 210 is the right answer. what answer were you getting herowise? i'm assuming you have the solutions, because how else do you know you're not getting it right etc. what's the answer?